method to find Limit of integral

Question

In a multiple choice question, the following is asked :

The limit $$\lim\limits_{x \to 0^+} \displaystyle\int_x^1 \left(1+ \dfrac 1t\right) e^{-\tfrac 1t} dt $$

is equal to :

1. $+\infty$
2. $-\infty$
3. $e^{-1}$
4. 0
5. 1

My choice went for (3) because the integral is convergent and the integrand is positive and - unless mistaken - it is bounded between $0$ and $\dfrac 2e$.

I would like to know whether what I did is correct and hopefully how to prove that the value of the limit is indeed $\dfrac 1e$

Answer 1

The antiderivative of $$(1+\frac1t)e^{-\frac1t}$$ is $$te^{-\frac1t} + C$$ (used integration by parts, you can differentiate to verify).

Evaluating this expression at $1$ gives $\frac1e$, the limit as $t\rightarrow0^+$ gives $0$, so the limit of the integral is just $\frac1e$, no need for L'Hopital's or any complex limit manipulation.

Answer 2

Yes, you're reasoning is correct. One need not evaluate the integral to determine the answer.

Since the integrand is positive and less than $1$, we can eliminate Choices $2$, $4$ and $5$.

And noting that $\lim_{t\to 0}\frac{e^{-1/t}}{t}=0$ goes to zero as $t\to 0$, the integral is finite and we can eliminate Choice $1$.

The only choice that remains is Choice $3$. And we are done!

Answer 3

By a change of variable, the integral is

$$\int_1^\infty \frac{1+u}{u^2}e^{-u}du.$$

As $$0<\frac{1+u}{u^2}e^{-u}< e^{1-u},$$

the integral is smaller than $1$, so 3.

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Also note that the integrand is the derivative of

$$-\frac{e^{-u}}u.$$